What is the sixtieth term in the following sequence? 1, 2, 4

I've learnt this method at school, and it can be applied easily to a wide variety of problems tht involve summations.

On observation we find that the 'n'th term of this sequence is of the form

\(T_n=T_{n-1}+ [n-1]\)

for n >= 2 i.e from the 2nd term onwards.

so to find the 60th term (\(T_60\)) all we need to do is this :

\(T_{60} = T_{59} + (60-1) = T_{59}+59\)
\(T_{59} = T_{58}+58\)
and
so
on
.
.
.
.
\(T_3 = T_2+2\)
\(T_2 = T_1+1\)
___________

On summing all the equations' left hand sides and right hand sides, observe that every equations' LHS (starting with the second) cancels with the term in the RHS above it - for example T59 in the second eqn in the LHS cancels with T59 in the RHS of the first equation and so on for the entire system of equations.

After this huge crash of dominoes, all we are left with is

\(T_{60} = T_1 + (1+2+3+4+...59)\)

\(1+2+...59\) as we know is \({59*(59+1)}/2 = 1770\)
and \(T_1 = 1\)

so the answer is 1771 (D)

Just my $0.02.